# Easy proof of trivial fusion implies normal p-complement

Theorem: Suppose G is a finite group with Sylow p-subgroup P. Then the following are equivalent:

• The set K of elements of G of order relatively prime to p (the p′-elements) form a subgroup
• If A and Ag are both subsets of P, then there is some x in CG(A) such that xg is in P

That the first implies the second is a silly trick: $(a^{-1} a^g)^{-1} = g^{-1} g^a \in P \cap K = 1$ for any $a \in P$ and $g \in K$ such that $a^g \in P$.

The second implies the first is not too hard (Frobenius normal p-complement theorem), but I'm trying to use this as a first example, and so don't want to have any prerequisites outside a very gentle undergraduate group theory course. Most of the rest of the talk is just using Sylow's theorem.

Is there a very low-tech, short, few-preliminaries proof that "absolutely no fusion" implies a normal p-complement?

I would be ok with assuming P is abelian, so that we get:

• If A and Ag are both subsets of P, then g in CG(A).

I'm also fine with assuming p = 2 so that "relatively prime to p" shortens to "odd".

I don't think using the transfer is appropriate, as it won't be used again, and the whole point of this proof is to motivate learning something else.

For a "no" answer to my question, it would be sufficient to convince me that transfer is needed (and nice if you can suggest a special case where it wouldn't be needed, other than P of order 2).

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You can do it with characters, but I don't suppose you would accept that as a "few preliminaries" proof. – Geoff Robinson Jun 18 '12 at 17:08
@Geoff: I know that no one in the target audience knows anything about characters, so it won't work for this. I do like that proof myself, and in general how characters and transfer can often accomplish the same tasks. One version is Isaacs's CT 8.22 (Brauer–Suzuki) books.google.com/books?id=MeE7BFXwQroC&pg=PA137 – Jack Schmidt Jun 18 '12 at 17:39
I am confused. Isn't $A_4$ a counterexample, with $P=V_4$, and $A$ being any subgroup of order 2? The set of odd order elements is certainly a group. Moreover, any $A^g$ is in $P$, since $P$ is normal. But $C_G(A)=1$, so there is no $x$ such that $xg\in P$ if $g$ is an element of order 3. – Alex B. Jun 18 '12 at 18:17
@Alex: (1,2,3) and (2,3,4) has a product of order 2, so the odd order elements are not a group. A4 satisfies both hypotheses with p=3, but neither with p=2. – Jack Schmidt Jun 18 '12 at 18:21
It's not hard to see that what you are asking for in the case $P$ abelian is equivalent to Burnside's Transfer Theorem. (If $A,A^g \subseteq P$ then $P$ and $P^{g^{-1}}$ are conjugate in $C_G(A)$, so $P^{hg}=P$ with $h \in C_G(A)$ and then the hypothesis of BTT hives $hg \in C_G(P)$, so $g \in C_G(A)$ and we have your hypothesis.) So you are really just asking for a transfer-free proof of BTT. – Derek Holt Jun 18 '12 at 18:48